Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a math problem, which I'm kind of lost in. I have the question:

Find the speed $ds/dt$ on the line $x=1+6t$, $y=2+3t$, $z=2t$. Integrate to find the length $s$ from $(1,2,0)$ to $(13,8,4)$. Check by using $12^{2} + 6^{2} + 4^{2}$.

I think I have found the speed: $\sqrt{6^{2}+3^{2}+2^{2}} = 7$. But im lost in integrating to finding the length between the two points. Can anyone help me?

David

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

If a particle travels along curve $s(t) = (1 + 6t, 2 + 3t, 2t)$ then its speed is $ds/ dt = (6, 3, 2)$.

The length of curve $s$ between two points $a = (1,2,0)$ and $b=(13, 8, 4)$ is computed as $\int_{t_a}^{t_b} \sqrt{(ds/dt)_x^2 + (ds/dt)_y^2 + (ds/dt)_z^2} dt$.

$t_a$ and $t_b$ are the points with the property $s(t_a) = a$ and $s(t_b) = b$. You need to determine them. Notice that $s(0) = a$ and $s(2) = b$.

With the values inserted, $$ \int_{t_a}^{t_b} \sqrt{6^2 + 3^2 + 2^2} dt= \int_0^2 \sqrt{49} dt = \int_0^2 7 dt = 14$$

share|improve this answer
    
Thanks man - helped a lot! –  user1090614 Mar 3 '13 at 16:39
    
@user1090614 You are welcome! –  goobie Mar 3 '13 at 16:41
add comment

Since this sounds like homework; hints:

  • speed is simply the magnitude of the velocity.
  • the magnitude of a vector $\vec{x}$ is $\sqrt{\vec{x} \cdot \vec{x}}$
  • given a time-dependent (or constant) expression for the speed, if you integrate it over a period, you'll find the length of the distance travelled.
  • i.e., you want an intermediate expression $s(t) = ...$, and in your case this expression will be very simple.
share|improve this answer
    
Am I correct when stating s(t) = 7. And therefore integrating s(t) between 0 and 2 = 14? –  user1090614 Mar 3 '13 at 16:31
    
sounds good to me... –  Eamon Nerbonne Mar 3 '13 at 16:35
    
Super - I totally forgot integrating over a constant c = xc+d... Sometimes the answer is right infront of you, but your eyes wont let you see it. –  user1090614 Mar 3 '13 at 16:37
    
Yeah, somehow particularly with math that's tricky. Wonder why... –  Eamon Nerbonne Mar 3 '13 at 16:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.